ON MULTIVALUED FIXED-POINT FREE MAPS ON Rn RAUSHAN Z. BUZYAKOVA Abstract. To formulate our results let f be a continuous multivalued map n from Rn to 2R and k a natural number such that |f (x)| ≤ k for all x. We prove that f is fixed-point free if and only if its continuous extension n f˜ : βRn → 2βR is fixed-point free. If one wishes to stay within metric terms, the result can be formulated as follows: f is fixed-point free if and n only if there exists a continuous fixed-point free extension f¯ : bRn → 2bR for some metric compactificaton bRn of Rn . Using the classical notion of colorablity, we prove that such an f is always colorable. Moreover, a number of colors sufficient to paint the graph can be expressed as a function of n and k only. The mentioned results also hold if the domain is replaced by any closed subspace of Rn without any changes in the range.

1. Introduction A series of topological results about fixed-point free maps are motivated by these two classical set-theoretical statements (see, in particular, [3]): S1. If f : X → X is a fixed-point free map, then there exists a finite cover F of X such that f (F ) misses F for each F ∈ F; and S2. Let P(X) be the set of all non-empty subsets of X and let f : X → P(X) be a map with the property that x 6∈ f (x). If there exists a natural number k such that |f (x)| ≤ k for all x ∈ X, then there exists a finite cover F of X such that F misses ∪{f (x) : x ∈ F } for each F ∈ F. One of the first results of topological nature related to these statements was obtained by Katetov in [6] by translating the first statement into this form: For a discrete space X, a map f : X → X is fixed-point free if and only if its continuous extension f˜ : βX → βX is fixed-point free. This topological version suggests that if one wants to have a similar criterion for a non-discrete space X one has to at least demand that elements of covers F in the statements under discussion be closed subsets of X. Thus, it is commonly accepted that when working with a topological space X and a continuous map f from a closed subspace A of X to X, any closed subset F of A that misses its image under f 1991 Mathematics Subject Classification. 54H25, 58C30, 54B20. Key words and phrases. fixed point, hyperspace, multivalued function. 1

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is called a color of f . If there exists a finite cover (”coloring”) of X by colors, f is said to be colorable. Katetov’s paper [6] and van Dowen’s work [4] have made a significant impact on topologists’ interest in the topic. A number of interesting results of topological nature in the direction of the first statement have been published since the mentioned papers (see, in particular, [8, Section 3.2] for references). In this paper we consider one of natural topological versions n of the second statement for Euclidean space Rn and its hyperspace 2R . In [2] it is proved that a continuous fixed-point free map from a closed subspace of Rn to Rn is colorable. In this paper we consider fixed-point free multivalued maps on Rn and its closed subspaces. To formulate our main results we first introduce the necessary terminology related to multivalued maps. For a topological space X, we use symbol 2X to denote the space of all nonempty closed subsets of X endowed with the Vietoris topology and symbol expk (X) to denote the subspace of 2X that consists of only those A ∈ 2X for which |A| ≤ k. A multivalued map f : X ⊂ Z → 2Z is fixed-point free if x 6∈ f (x) for every x ∈ X. A closed set F ⊂ X is a color of a continuous map f from X ⊂ Z to 2Z if F misses ∪{f (x) : x ∈ F }. If X can be covered by finitely many colors of f then f is said to be colorable. The main results of the paper are Theorems 2.7 and 2.8, which state that any continuous fixed-point free map from a closed subspace of Rn to expk (Rn ) is colorable and there exists a formula that computes a number of colors sufficient for painting in terms of n and k only. Using the main result we also show that a criterion similar to the Katetov’s holds for multivalued maps as well. Namely, we show that a continuous map f from a closed subspace X of Rn to expk (Rn ) is fixed-point free if and only if its continuous extension f˜ : βX → expk (βRn ) is fixed-point free. To justify the requirement on sizes of f (x) in our main results let us consider one simple example. Define f from ω \ {0} to the space of finite subsets of ω as follows: f (n) = {n + 1, n + 2, ..., 2n}. The map f is continuous because ω and the space of finite subsets of ω are discrete. Since for every n ∈ ω \ {0}, all elements of {n + 1, n + 2, ..., 2n} must be of different color we conclude that f is not colorable. This example justifies our requirement in the main results that the set {|f (x)| : x ∈ X} is bounded by a positive integer. Before we dive into the technical part of the paper we would like to outline a short transparent argument of the main results of the paper for a fixed-point free map f : R → exp2 (R). For this let f1 (x) = min f (x) and f2 (x) = max f (x). Since f is fixed-point free, the maps f1 and f2 are fixed-point free real-valued maps. By [1, Theorem 2.5], both functions are colorable. If one lets F1 and F2 be colorings of these maps then it is easy to verify that the family {A ∩ B : A ⊂ F1 , B ⊂ F2 } is a coloring of f . If one wishes to extend the argument for the case exp3 (R) using a straightforward inductive approach, then one may find oneself dealing with open colors or with a single-valued map with the domain being a proper subset of the

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range. Existence of open colors in a single-valued case can be easily deduced from the definition of colorability when one deals with self-maps. However, if one works with maps from a subspace X of Y into Y , a work needs to be done. Nevertheless, the presented argument can be extended for expk (R) for any k with some more work and suitable references. Although our argument for any n and k that we present in the paper may seem different from the one just described, a closer look will reveal that it carries the same idea hidden behind technical details naturally arising when dealing with higher dimensions. Throughout the paper we will follow standard notation and terminology as in [5]. 2. Results For simplicity, but without loss of generality, most of our arguments related to Rn will be carried out for R5 . This will free the letter ”n” for other purposes. Since throughout our discussion we will juggle several spaces at a time we agree that by S¯ we denote the closure of S in Rk (where the value of k is always understood from the context) while clX (P ) will denote the closure of P in X. A standard neighborhood in 2X will be denoted as hU1 , ..., Um i = {A ∈ 2X : A ⊂ U1 ∪ ... ∪ Um and Ui ∩ A 6= ∅ f or all i = 1, ..., m}, where U1 , ..., Um are open sets of X. By expn (X) we denote the subspace {F ∈ 2X : |F | ≤ n}. Definition 2.1. Let f be a continuous map from X ⊂ Z to 2Z . A closed set F ⊂ X is a bright color of f if F misses clZ [∪{f (x) : x ∈ F }].

Proposition 2.2. Suppose Z is normal, X is closed in Z, f : X → 2Z is continuous, and F is a bright color of f . Then there exists an open neighborhood U of F such that clX (U ) is a bright color of f . Proof. Since F is a bright color of f and X is closed in Z, the sets F and clZ (∪{f (x) : x ∈ F }) are disjoint closed sets in Z. Since Z is normal, there exist V an open neighborhood of clZ [∪{f (x) : x ∈ F }] in Z and W an open neighborhood of F in Z such that clZ (W ) misses clZ (V ). Consider the open set hV i in 2Z . Since f is continuous and f (F ) ⊂ hV i there exists an open neighborhood U of F in X such that U ⊂ W and f (clZ (U )) ⊂ hV i. Thus, clZ (∪{f (x) : x ∈ clZ (U )}) is in clZ (V ). The latter misses clZ (W ) and therefore clZ (U ). Therefore, U is as desired. 

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Proposition 2.2 implies that if X is closed in Rk and F is an m-sized bright coloring of a continuous map f : X → expn (Rk ) then there exists an m-sized open cover U of X the closures of whose elements are bright colors of f . We will use this observation throughout the paper without formally referencing it. In the following discussion that leads to the main result we will restrict ourselves to closed subspace of R5 . This is done to avoid accumulation of too many variables. All arguments are valid if one replace ”5” with any natural number. Definition 2.3. S(5, n) denotes the following statement: there exists the smallest integer K(5, n) such that every continuous fixed-point free map f from a closed subset X ⊂ R5 to expn (R5 ) is colorable by at most K(5, n) many bright colors. Lemma 2.4. Suppose f, g, h : X ⊂ Z → 2Z are maps; and g and h are colorable by at most Ng and Nh (bright) colors, respectively. If f (x) ⊂ g(x) ∪ h(x) for every x ∈ X then f is colorable by at most Ng · Nh (bright) colors. Proof. Let G and H be bright colorings of g and h of sizes Ng and Nh , respectively. Put F = {G ∩ H : G ∩ H 6= ∅, H ∈ H, G ∈ G}. Clearly, |F| ≤ Ng · Nh . Since G and H are closed covers of X, so is F. Fix F = H ∩ G in F. Since, by hypothesis, f (x) ⊂ g(x) ∪ h(x), we conclude that clZ [∪{f (x) : x ∈ H ∩G}] ⊂ clZ [∪{g(x) : x ∈ H ∩G}]∪clZ [∪{h(x) : x ∈ H ∩G}]. Since G and H are bright colorings, clZ [∪{g(x) : x ∈ H ∩ G}] and clZ [∪{h(x) : x ∈ H ∩ G}] miss H ∩ G. Therefore, the left side of the above set inclusion formula misses H ∩ G as well, meaning H ∩ G is a bright color for f .  In what follows, by πi we denote the projection of R5 onto its i-th coordinate axis. The next two statements (Lemmas 2.5 and 2.6) hold if we replace π1 by πi for any i ∈ {1, ..., n}. However, we will carry out our arguments for π1 for the already mentioned reason of avoiding unnecessary variables. Lemma 2.5. Suppose n > M ≥ 1; A is closed in R5 ; f : A → expn (R5 ) \ expn−1 (R5 ) is continuous and fixed-point free; |{y ∈ f (x) : π1 (y) = max π1 (f (x))}| = M for all x ∈ A; and S(5, n − 1) is true. Then f is colorable by at most [K(5, n − 1)]2 bright colors. Proof. Define g and h from A to expn−1 (R5 ) as follows: g(x) = {z ∈ f (x) : π1 (z) = max π1 (f (x))} and h(x) = f (x) \ g(x). Since f (x) is finite, max π1 (f (x)) exists. Hence g(x) is defined. Since |{y ∈ f (x) : π1 (y) = max π1 (f (x))}| = M and 1 ≤ M < n, we conclude that 0 <

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|f (x) \ g(x)| < n and 0 < |g(x)| < n. Therefore, g and h are well defined functions from A to expn−1 (R5 ). Observe that f (x) = g(x) ∪ h(x) for each x. Therefore, by Lemma 2.4, to reach the conclusion of our lemma we need to show that g and h are colorable by at most K(5, n − 1) bright colors each. Since we assume that S(5, n−1) holds it suffices to show that g and h are continuous and fixed-point free. The latter property follows from the facts that f is fixed-point free and f (x) = g(x) ∪ h(x). To prove continuity of g and h, fix x ∈ A and open neighborhoods Ug(x) and Vh(x) of g(x) and h(x) in expn−1 (R5 ). We need to find an open neighborhood of x in A whose image under g(x) and h(x) are in Ug(x) and Vh(x) , respectively. Without loss of generality we may assume that the selected neighborhoods are standard, that is, in the form Ug(x) = hUy : y ∈ g(x)i and Vh(x) = hUy : y ∈ f (x) \ g(x)i, where Uy is a fixed bounded open neighborhood of y in R5 for each y ∈ f (x). We may also assume that the following properties hold. P1: min π1 (U y ) > max π1 (U z ) whenever y ∈ g(x) and z ∈ f (x) \ g(x). P2: U y ∩ U z = ∅ for any distinct y, z ∈ f (x). The property P1 can be achieved since by the definition of g, the set π1 (g(x)) is a singleton and its element is strictly greater than any element of π1 (f (x)\g(x)). Put Wf (x) = hUy : y ∈ f (x)i. Clearly, Wf (x) is an open neighborhood of f (x). By continuity of f , there exists an open O of x in A such that f (O) ⊂ Wf (x) . To finish the proof of continuity of g and h it suffices to show that g(O) ⊂ Ug(x) and h(O) ⊂ Vh(x) . For this fix an arbitrary x0 ∈ O. By the choice of O, we have f (x0 ) ∈ Wf (x) . Let f (x0 ) = {z1 , ..., zn } ∈ expn (R5 ) \ expn−1 (R5 ) and π1 (zi ) = max π1 (f (x0 )) for any i = 1, ..., M . By the lemma’s condition on M , we have π1 (zj ) < max π1 (f (x)) for any j = M + 1, ..., n. By P1, we have P3: zi ∈ ∪{Uy : y ∈ g(x)} for any i = 1, ..., M . By P2 and P3, we have P4: zj ∈ ∪{Uy : y ∈ f (x) \ g(x)} for any j = M + 1, ..., n. By P2 and the fact that f (x0 ) ∈ Wf (x) = hUy : y ∈ f (x)i, we conclude that each Uy , participating in the definition of Wf (x) , contains exactly one zi ∈ f (x0 ). By P3 and P4, we have {z1 , ..., zM } ∈ Ug(x) and {zM +1 , ..., zn } ∈ Vh(x) . Since g(x0 ) = {z1 , ..., zM } and h(x0 ) = {zM +1 , ..., zn }, we are done. 

Lemma 2.6. Suppose n > 1; A is closed in R5 ; f : A → expn (R5 ) \ expn−1 (R5 ) is continuous and fixed-point free; |{y ∈ f (x) : π1 (y) = max π1 (f (x))}| > 1 for all x ∈ A; and S(5, n−1) is true. Then f is colorable by at most n·[K(5, n−1)]2 bright colors.

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Proof. For m = 1, ..., n − 1, define Om ⊂ A as follows: x ∈ Om if and only if Mx = |{y ∈ f (x) : π1 (y) = max π1 (f (x))}| < n − m. Claim. Om is open. To prove the claim, fix x ∈ Om and let Vf (x) = hVy : y ∈ f (x)i be an open neighborhood of f (x) such that the following hold: (1) Vy is a bounded neighborhood of y for every y ∈ f (x); (2) Vy ∩ Vz = ∅ for any distinct y, z ∈ f (x); and (3) min π1 (V y ) > max π1 (V z ) whenever π1 (y) = max π1 (f (x)) and π1 (z) < max π1 (f (x)). It suffices to show now that f −1 (Vf (x) ) ⊂ Om . For this pick x0 ∈ f −1 (Vf (x) ). We have f (x0 ) ∈ Vf (x) . By (2) and (3), Mx0 ≤ Mx . Hence, Mx0 < n − m. By the definition of Om , x0 ∈ Om . The claim is proved. We will construct our coloring inductively. For short put K = [K(5, n − 1)]2 . Step 1. Put A1 = A \ O1 . Thus, an element x of A is in A1 if and only if Mx ≥ n − 1. Since |π1 (f (x))| > 1 for every x ∈ A we conclude that Mx = n − 1 for every x ∈ A1 . Therefore, by Lemma 2.5, there exists a K-sized open cover U1 of A1 the closures of whose elements are bright colors for f . Assumption. Assume for m − 1 an open family Um−1 of size at most (m − 1) · K is defined and the following conditions are met: P1: U is a bright color of f for every S U ∈ Um−1 ; and P2: If Mx ≥ n − (m − 1) then x ∈ Um−1 . S Step m < n. Put Am = A \ [Om ∪ ( Um−1 )]. By Claim, the set Am isSclosed. Pick any x ∈ Am . Then x 6∈ Om , meaning that Mx ≥ n − m. Also x 6∈ Um−1 , which, by P2, implies Mx < n − (m − 1). Thus, n − m ≤ Mx < n − m + 1. Therefore, Mx = n − m. Therefore, by Lemma 2.5, there exists a K-sized open cover Vm of Am the closures of whose elements are bright colors for f . Put Um = Um−1 ∪ Vm . Clearly, the size of this family is at most m · K. Let us verify P1 and P2 for m. Property P1 holds since Um is the union of two families satisfying P1. For P2, observe that Mx ≥ n − m if and only if x 6∈ Om . But Um covers the complement of Om . S The construction is complete. It suffices to show now that A = Un−1 . For this S pick x ∈ A. By the lemma’s hypothesis, Mx > 1 = n − (n − 1). By P2, x ∈ Un−1 .  The base step in the proof of our main theorem uses the following statement ([2, Proposition 2.9]): If f is a continuous fixed-point free map from a closed subset X of Rm to Rm then f is colorable by at most m + 3 bright colors.

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Theorem 2.7. There exists an integer K(m, n) such that every continuous fixed-point free map from a closed subspace X of Rm into expn (Rm ) is colorable by at most K(m, n) many bright colors. Proof. As in previous statements, to avoid accumulation of extra variables, let us deal with m = 5. Thus, we need to show that K(5, n) exists for every natural number n. By [2, Proposition 2.9], K(5, 1) exists. Assume that K(5, n − 1) exists. To prove the conclusion of the theorem for n, fix any fixed-point free continuous map f from a closed subspace X of R5 into expn (R5 ). Next define L ⊂ X as follows: x ∈ L if and only if |f (x)| < n. Then L is closed and the range of f |L is a subset of expn−1 (R5 ) (notice that L can be empty). Therefore, by our inductive assumption, there exists a K(5, n − 1)-sized open S cover UL of L the closures of whose elements are bright colors. Put E = X \ UL . Then E is closed and |f (x)| = n for every x ∈ E. For 1 ≤ i ≤ n, define Si as follows: x ∈ Si if and only if x ∈ E and |πi (f (x))| = 1. Notice that Si can be empty. Clearly, Si is closed. Inductively, we will first cover ∪i≤n Si = {x ∈ E : |πi (f (x))| = 1 f or some i} by bright colors and then we will cover the rest of E. Step 1. Put Ei = ∩j6=i Sj . Notice that Ei can be empty. Since |f (x)| = n for every x ∈ E we conclude that |πi (f (x))| = n for all x ∈ Ei . Since n > 1, by Lemma 2.5 (with π1 replaced by πi and M = 1), there exists a finite open cover U1 of ∪i≤n Ei the closures of whose elements are bright colors for f . Assumption. Suppose an open finite family Uk−1 , where 1 ≤ k − 1 < n, is defined and the following hold: P1: For every S at most (k − 1)-sized subset I of {1, ..., n} the inclusion ∩j6∈I Sj ⊂ Uk−1 holds; P2: U is a bright color for every U ∈ Uk−1 . S Step k < n. For every k-sized I ⊂ {1, ..., n} put EIS= [∩j6∈I Sj ] \ [ Uk−1 ]. Pick i∗ ∈ I. Then |I \ {i∗ }| =Sk − 1. By P1, the set Uk−1 contains ∩{Sj : j ∈ I \{i∗ }}. Since EI misses Uk−1 , we conclude that |πi∗ (f (x))| > 1 for every x in EI . By Lemma 2.6, there exists a finite open cover UI of EI the closures of whose elements are bright colors. Put Uk = [∪{UI : I ⊂ {1, ..., n}, |I| = k}] ∪ Uk−1 Properties P1 and P2 clearly hold for k. The construction is complete. Let us show that {x ∈ E : |πi (f (x))| = 1 f or some i} is covered by Un−1 . For this pick any z in this set. Put Iz = {i ≤ n : |πi (f (x))| > 1}. Clearly, |Iz | < n. Since x ∈ E we conclude S that |f (x)| = n. Therefore, Iz 6= ∅. Therefore z ∈ ∩j6∈Iz Sj . By P1, z ∈ Un−1 . S To finish the proof we need to cover E \ [ Un−1 ] by bright colors. For this S observe that E \ [ Un−1 ] contains only those x for which |πi (f (x))| = n > 1

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for all i. Therefore, by Lemma 2.5 (with M = 1), the set in question is covered by bright colors. Since we always used Lemmas 2.5 and 2.6 to construct our coloring, the size of the coloring depends on n and number 5 only. Thus, K(5, n) exists.  A non-technical version of the above result is the following theorem, where n and k denote any natural numbers. Theorem 2.8. Any continuous fixed-point free map from a closed subspace X of Rk into expn (Rk ) is brightly colorable. Observe that if a continuous map f : X → 2Z has the property that f (x) is compact in Z for all x then there exists the continuous extension f˜ : βX → 2βZ . For our next discussion put K(X) = {A ∈ 2X : A is compact}. In [2] it is proved that a continuous map f from a closed subspace X of Rk to Rk is fixed-point free if and only if its continuous extension f˜ : βX → βRk is fixed-point free. It is natural to ask if the corresponding statement holds for a multivalued map f : X → expn (Rk ) and its continuous extension f˜ : βX → expn (βRk ). Observe first that the continuous extension exists since expn (βRk ) is compact. The affirmative answer to this question follows from the next statement. Proposition 2.9. Let X be a closed subspace of a normal space Z. If F is a bright coloring of a continuous map f : X → K(Z), then {βF : F ∈ F} is a bright coloring of f˜ : βX → K(βZ). Proof. Since X is normal and F is a finite closed cover of X the family {βF : F ∈ F} is a closed cover of βX. Therefore, we only need to show that βF is a bright color for f˜. Since F is a bright color for f the set F misses clZ (∪{f (x) : x ∈ F }). Since F and clZ (∪{f (x) : x ∈ F }) are disjoint closed subsets of the normal space Z we conclude that βF misses clβZ (∪{f (x) : x ∈ F }). Since f˜ is the continuous extension of f we conclude that clβZ (∪{f˜(x) : x ∈ βF }) = clβZ (∪{f (x) : x ∈ F }). Thus βF misses clβZ (∪{f˜(x) : x ∈ βF }), whence βF is a bright color of f˜. 

Theorem 2.10. Let f be a continuous map from a closed subspace X of Rk to expn (Rk ). Then f is fixed-point free if and only if f˜ : βX → expn (βRk ) is fixed-point free.

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Proof. Sufficiency is obvious. To prove necessity, let F be a bright coloring of f . By Proposition 2.9, {βF : F ∈ F} is a coloring for f˜. Since {βF : F ∈ F} covers βX and f (βF ) misses βF for each F ∈ F, we conclude that f˜ does not fix any point.  Using spectral techniques it is observed in [8, Corollary 3.5.7] that if f is a continuous colorable self-map on a separable metric space X then one can find a continuous fixed-point free extension f¯ : bX → bX, where bX is a metric compactification of X of the same dimensionality as that of X. Using the same technique we will next outline a proof for the following metric version of Theorem 2.10 for natural numbers n and k. Theorem 2.11. Let f : Rk → expn (Rk ) be continuous. Then f is fixed-point free if and only if there exists a continuous fixed-point free extension f¯ : bRk → expn (bRk ), where bRk is some metric compactification of Rk of dimension k. Proof. Sufficiency is obvious. Let us outline a proof of necessity. By Theorem k ˇepin spectral 2.10, f˜ : βRk → expn (βRk ) ⊂ 2βR is fixed-point free. By Sˇ k

theorem [7], we can find spectra {bα (Rk ), παγ , A} and {2bα (R ) , pγα , A} with inverse k limits βRk and 2βR , respectively, and a family of maps {fα : α ∈ A} such that the following hold: (1) bα (Rk ) is a metric compactification of Rk of dimension k for all α k (2) παγ and pγα are identity maps on Rk and 2R , respectively. (3) fα ◦ πα = pα ◦ f . Since f˜ is fixed-point free and bα (Rk ) is compact for every α we may assume that fα is fixed-point free for every α. By (2) and (3), each fα coincides with f on Rk . Therefore, each {fα , bα (Rk )} serves our purpose.  We would like to finish the paper by commenting on a number of colors sufficient to paint a given graph. If one follows the argument of Theorem 2.7 or the argument for the reals outlined in the introduction one will quickly see that the size of the coloring constructed for the case expn (Rk ) is at least k n . Therefore, it is natural to wonder if an estimate for the required number of colors can be represented as a polynomial of both n and k. References [1] R. Buzyakova, Fixed Point Free Maps on the Reals and More, Top. and its Appl., 156 (2008), 465-472. [2] R. Buzyakova and A. Chigogidze, Fixed-point free maps of Euclidean spaces, Fundamenta Mathematica, 212 (2011), 1-16.

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[3] N. G. de Bruijn and P. Erd¨os, A color problem for infinite graphs and a problem in the theory of relations, Indagationes Math., 13 (1951), 371-373. [4] E. van Douwen, βX and fixed-point free maps, Topology Appl. 51 (1993), 191–195. [5] R. Engelking, General Topology, PWN, Warszawa, 1977. [6] M. Katˇetov, A theorem on mappings, Comm. Math. Univ. Carolinae 8 (1967), 431–433. ˇcepin, Topology of limit spaces of uncountable inverse spectra, (Russian), Uspekhi [7] E. V. Sˇ Mat. Nauk, 31 (1976), 191-226. [8] J. van Mill, The infinite-dimensional topology of function spaces, Elsevier, Amsterdam, 2001. Department of Mathematics and Statistics, The University of North Carolina at Greensboro, Greensboro, NC, 27402, USA E-mail address: [email protected]